已知数列{an}其前n项和为Sn=3/2n^2+7/2n(n属于正整数)
问题描述：
已知数列{an}其前n项和为Sn=3/2n^2+7/2n(n属于正整数)
设cn=9/(2an-7)(2an-1),数列{cn}的前n项和为Tn,求使不等式Tn>k/57对一切n属于正整数都成立的最大整数k的值
答
Sn=(3/2)n^2+(7/2)n
n=1,a1=5
an =Sn - S(n-1)
= (3/2)(2n-1) +7/2
= 3n+2
cn=9/[(2an-7)(2an-1)]
=9/[(6n-3)(6n+3)]
= 1/[(2n-1)(2n+1)] >0
Tn =c1+c2+...+cn
≥T1
= 1/3
1/3 > k/57
k = 57/3 = 19= 1/[(2n-1)(2n+1)] >0是设的吗2n-1 >0; n=1,2,3,...2n+1 >0 ; n=1,2,3,....=> 1/[(2n-1)(2n+1)] >0